Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
MEM2(g2(x, y), z) -> MEM2(x, z)
F2(x, g2(y, z)) -> F2(x, y)
++12(x, g2(y, z)) -> ++12(x, y)
MEM2(x, max1(x)) -> NULL1(x)
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
MEM2(g2(x, y), z) -> MEM2(x, z)
F2(x, g2(y, z)) -> F2(x, y)
++12(x, g2(y, z)) -> ++12(x, y)
MEM2(x, max1(x)) -> NULL1(x)
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MAX1(g2(g2(g2(x, y), z), u)) -> MAX1(g2(g2(x, y), z))
Used argument filtering: MAX1(x1) = x1
g2(x1, x2) = g1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MEM2(g2(x, y), z) -> MEM2(x, z)
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
MEM2(g2(x, y), z) -> MEM2(x, z)
Used argument filtering: MEM2(x1, x2) = x1
g2(x1, x2) = g1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
++12(x, g2(y, z)) -> ++12(x, y)
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
++12(x, g2(y, z)) -> ++12(x, y)
Used argument filtering: ++12(x1, x2) = x2
g2(x1, x2) = g1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F2(x, g2(y, z)) -> F2(x, y)
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F2(x, g2(y, z)) -> F2(x, y)
Used argument filtering: F2(x1, x2) = x2
g2(x1, x2) = g1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
++2(x, nil) -> x
++2(x, g2(y, z)) -> g2(++2(x, y), z)
null1(nil) -> true
null1(g2(x, y)) -> false
mem2(nil, y) -> false
mem2(g2(x, y), z) -> or2(=2(y, z), mem2(x, z))
mem2(x, max1(x)) -> not1(null1(x))
max1(g2(g2(nil, x), y)) -> max'2(x, y)
max1(g2(g2(g2(x, y), z), u)) -> max'2(max1(g2(g2(x, y), z)), u)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.